The incircle of triangle $ABC$ touches $BC$ at $D$ and $AC$ at $E$. Let $K$ be the point on $CB$ with $CK=BD$, and $L$ be the point on $CA$ with $AE=CL$. Lines $AK$ and $BL$ meet at $P$. If $Q$ is the midpoint of $BC$, $I$ the incenter, and $G$ the centroid of $\triangle ABC$, show that: $(a)$ $IQ$ and $AK$ are parallel, $(b)$ the triangles $AIG$ and $QPG$ have equal area.

A cyclic quadrilateral $ABCD$ has circumcircle $\Gamma$, and $AB+BC=AD+DC$. Let $E$ be the midpoint of arc $BCD$, and $F (\neq C)$ be the antipode of $A$ [i]wrt[/i] $\Gamma$. Let $I,J,K$ be the incenter of $\triangle ABC$, the $A$-excenter of $\triangle ABC$, the incenter of $\triangle BCD$, respectively. Suppose that a point $P$ satisfies $\triangle BIC \stackrel{+}{\sim} \triangle KPJ$. Prove that $EK$ and $PF$ intersect on $\Gamma.$

Let $BE$ and $CF$ be the bisectors of $\angle B$ and $\angle C$ of a triangle $ABC$ whose incentre is $I$. Suppose $EF$, extended, meets the circumcircle of $ABC$ in $M,N$. Show that the circumradius of $MIN$ is twice that of $ABC$.

In the convex pentagon $ABCDE$ all interior angles have the same measure. Prove that the perpendicular bisector of segment $EA$, the perpendicular bisector of segment $BC$ and the angle bisector of $\angle CDE$ intersect in one point.

Given a parabola $C: y^2=4px\ (p>0)$ with focus $F(p,\ 0)$. Let two lines $l_1,\ l_2$ passing through $F$ intersect orthogonaly each other, $C$ intersects with $l_1$ at two points $P_1,\ P_2$ and $C$ intersects with $l_2$ at two points $Q_1,\ Q_2$. Answer the following questions. (1) Set the equation of $l_1$ as $x=ay+p$ and let the coordinates of $P_1,\ P_2$ as $(x_1,\ y_1),\ (x_2,\ y_2)$, respectively. Express $y_1+y_2,\ y_1y_2$ in terms of $a,\ p$. (2) Show that $\frac{1}{P_1P_2}+\frac{1}{Q_1Q_2}$ is constant regardless of way of taking $l_1,\ l_2$.

A half-circle of diameter $EF$ is placed on the side $BC$ of a triangle $ABC$ and it is tangent to the sides $AB$ and $AC$ in the points $Q$ and $P$ respectively. Prove that the intersection point $K$ between the lines $EP$ and $FQ$ lies on the altitude from $A$ of the triangle $ABC$. [i]Albania[/i]

In a triangle $ABC$ the point $D$ is the intersection of the interior angle bisector of $\angle BAC$ and side $BC$. Let $P$ be the second intersection point of the exterior angle bisector of $\angle BAC$ with the circumcircle of $\angle ABC$. A circle through $A$ and $P$ intersects line segment $BP$ internally in $E$ and line segment $CP$ internally in $F$. Prove that $\angle DEP = \angle DFP$.

Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$. (Here we always assume that an angle bisector is a ray.) [i]Proposed by Sergey Berlov, Russia[/i]

Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.

Let $ABC$ be an equilateral triangle. $M$ is the midpoint of segment $AB$ and $N$ is the midpoint of segment $BC$. Let $P$ be the point outside $ABC$ such that the triangle $ACP$ is isosceles and right in $P$. $PM$ and $AN$ are cut in $I$. Prove that $CI$ is the bisector of the angle $MCA$ .

Let $ABC$ be a triangle with area $1$ and $P$ the middle of the side $[BC]$. $M$ and $N$ are two points of $[AB]-\left \{ A,B \right \} $ and $[AC]-\left \{ A,C \right \}$ respectively such that $AM=2MB$ and$CN=2AN$. The two lines $(AP)$ and $(MN)$ intersect in a point $D$. Find the area of the triangle $ADN$.

In the acute-angled triangle $ABC$, with $AB \ne AC$, $D$ is the foot of the angle bisector of angle $A$, and $E, F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$. Leonard Giugiuc

Does a parallelogram exist such that all pairwise meets of bisectors of its angles are situated outside it?

In triangle $ ABC$, $ AB \equal{} 13$, $ BC \equal{} 14$, and $ AC \equal{} 15$. Let $ D$ denote the midpoint of $ \overline{BC}$ and let $ E$ denote the intersection of $ \overline{BC}$ with the bisector of angle $ BAC$. Which of the following is closest to the area of the triangle $ ADE$? $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.5 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 3.5 \qquad \textbf{(E)}\ 4$

In a non-isosceles acute triangle $ABC$, $D$ is the midpoint of the edge $[BC]$. The points $E$ and $F$ lie on $[AC]$ and $[AB]$, respectively, and the circumcircles of $CDE$ and $AEF$ intersect in $P$ on $[AD]$. The angle bisector from $P$ in triangle $EFP$ intersects $EF$ in $Q$. Prove that the tangent line to the circumcirle of $AQP$ at $A$ is perpendicular to $BC$.

In an acute triangle $ABC,$ let $D$ be a point on the side $BC.$ Let $M_1, M_2, M_3, M_4, M_5$ be the midpoints of the line segments $AD, AB, AC, BD, CD,$ respectively and $O_1, O_2, O_3, O_4$ be the circumcenters of triangles $ABD, ACD, M_1M_2M_4, M_1M_3M_5,$ respectively. If $S$ and $T$ are midpoints of the line segments $AO_1$ and $AO_2,$ respectively, prove that $SO_3O_4T$ is an isosceles trapezoid.

Let $ABC$ be a triangle with $AB < AC$ and incentre $I$. Let $E$ be the point on the side $AC$ such that $AE = AB$. Let $G$ be the point on the line $EI$ such that $\angle IBG = \angle CBA$ and such that $E$ and $G$ lie on opposite sides of $I$. Prove that the line $AI$, the line perpendicular to $AE$ at $E$, and the bisector of the angle $\angle BGI$ are concurrent.

$ABC$ is a triangle $E, F$ are points in $AB$, such that $AE = EF = FB$ $D$ is a point at the line $BC$ such that $ED$ is perpendiculat to $BC$ $AD$ is perpendicular to $CF$. The angle CFA is the triple of angle BDF. ($3\angle BDF = \angle CFA$) Determine the ratio $\frac{DB}{DC}$. %Edited!%

In the triangle $ABC$, let $AM$ and $CN$ internal bisectors, with $M$ in $BC$ and $N$ in $AB$. Prove that if $$\frac{\angle BNM}{\angle MNC}=\frac{\angle BMN}{\angle NMA}$$ then $ABC$ is isosceles.

Points $X$ and $Y$ are the midpoints of arcs $AB$ and $BC$ of the circumscribed circle of triangle $ABC$. Point $T$ lies on side $AC$. It turned out that the bisectors of the angles $ATB$ and $BTC$ pass through points $X$ and $Y$ respectively. What angle $B$ can be in triangle $ABC$?

The sides $ a,b$ and the bisector of the included angle $ \gamma$ of a triangle are given. Determine necessary and sufficient conditions for such triangles to be constructible and show how to reconstruct the triangle.

In the triangle $ABC$, angle bisector $CD$ intersects the circumcircle of $ABC$ at the point $K$. (a) Prove the equalities: $$\frac1{ID}-\frac1{IK}=\frac1{CI},\enspace\frac{CI}{ID}-\frac{ID}{DK}=1$$where $I$ is the center of the inscribed circle of triangle $ABC$. (b) On the segment $CK$ some point $P$ is chosen whose projections on $AC,BC,AB$ respectively are $P_1,P_2,P_3$. The lines $PP_3$ and $P_1P_2$ intersect at a point $M$. Find the locus of $M$ when $P$ moves around segment $CK$.

Let $ ABC$ be a triangle. The $ A$-excircle of triangle $ ABC$ has center $ O_{a}$ and touches the side $ BC$ at the point $ A_{a}$. The $ B$-excircle of triangle $ ABC$ touches its sidelines $ AB$ and $ BC$ at the points $ C_{b}$ and $ A_{b}$. The $ C$-excircle of triangle $ ABC$ touches its sidelines $ BC$ and $ CA$ at the points $ A_{c}$ and $ B_{c}$. The lines $ C_{b}A_{b}$ and $ A_{c}B_{c}$ intersect each other at some point $ X$. Prove that the quadrilateral $ AO_{a}A_{a}X$ is a parallelogram. [i]Remark.[/i] The $ A$[i]-excircle[/i] of a triangle $ ABC$ is defined as the circle which touches the segment $ BC$ and the extensions of the segments $ CA$ and $ AB$ beyound the points $ C$ and $ B$, respectively. The center of this circle is the point of intersection of the interior angle bisector of the angle $ CAB$ and the exterior angle bisectors of the angles $ ABC$ and $ BCA$. Similarly, the $ B$-excircle and the $ C$-excircle of triangle $ ABC$ are defined. [hide="Source of the problem"][i]Source of the problem:[/i] Theorem (88) in: John Sturgeon Mackay, [i]The Triangle and its Six Scribed Circles[/i], Proceedings of the Edinburgh Mathematical Society 1 (1883), pages 4-128 and drawings at the end of the volume.[/hide]

In triangle $ABC$ $CL$ is a bisector($L$ lies $AB$) $I$ is center incircle of $ABC$. $G$ is intersection medians. If $a=BC, b=AC, c=AB$ and $CL\perp GI$ then prove that $\frac{a+b+c}{3}=\frac{2ab}{a+b}$

Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.